Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10000 + 10;
int gauss_elim(int *t, int n, vector<pair<int, int> > &v) {
  v.clear();
  int j = 0;
  for (int i = 0; i < 31; i++) {
    int x = -1;
    for (int k = j; k < n; k++)
      if (t[k] & (1 << i)) {
        x = k;
        break;
      }
    if (x == -1) continue;
    if (x != j) {
      swap(t[x], t[j]);
      v.push_back(pair<int, int>(x, j));
      v.push_back(pair<int, int>(j, x));
      v.push_back(pair<int, int>(x, j));
    }
    for (int k = j + 1; k < n; k++)
      if (t[k] & (1 << i)) v.push_back(pair<int, int>(k, j)), t[k] ^= t[j];
    j++;
  }
  return j;
}
int tmp[50], tmp2[50];
bool decomp(int *t, int n, int val, int *res) {
  memset(tmp2, 0, sizeof(tmp2));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 30; j++)
      if (t[i] & (1 << j)) tmp2[j] |= (1 << i);
  for (int j = 0; j < 30; j++)
    if (val & (1 << j)) tmp2[j] |= (1 << n);
  vector<pair<int, int> > nouse;
  gauss_elim(tmp2, 30, nouse);
  fill(res, res + n, 0);
  for (int i = 29; i >= 0; i--)
    if (tmp2[i]) {
      if (tmp2[i] == (1 << n)) return 0;
      int j = 0;
      while (!(tmp2[i] & (1 << j))) j++;
      res[j] = (tmp2[i] & (1 << n)) ? 1 : 0;
      for (int k = j + 1; k < n; k++)
        if (tmp2[i] & (1 << k)) res[j] ^= res[k];
    }
  return 1;
}
int a[maxn], b[maxn];
vector<pair<int, int> > v1, v2;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  int x = gauss_elim(a, n, v1);
  int y = gauss_elim(b, n, v2);
  reverse(v2.begin(), v2.end());
  for (int i = 0; i < y; i++) {
    if (!decomp(a, x, b[i], tmp)) {
      printf("-1\n");
      return 0;
    }
    if (!tmp[i])
      for (int j = i + 1; j < x; j++)
        if (tmp[j]) {
          swap(tmp[i], tmp[j]);
          swap(a[i], a[j]);
          v1.push_back(pair<int, int>(i, j));
          v1.push_back(pair<int, int>(j, i));
          v1.push_back(pair<int, int>(i, j));
        }
    if (!tmp[i]) {
      printf("FUCK 1\n");
      return 0;
    }
    tmp[i] = 0;
    for (int j = 0; j < x; j++)
      if (tmp[j]) v1.push_back(pair<int, int>(i, j)), a[i] ^= a[j];
    if (a[i] != b[i]) {
      printf("FUCK 2\n");
      return 0;
    }
  }
  for (int i = y; i < x; i++) v1.push_back(pair<int, int>(i, i));
  printf("%d\n", v1.size() + v2.size());
  for (auto i : v1) printf("%d %d\n", i.first + 1, i.second + 1);
  for (auto i : v2) printf("%d %d\n", i.first + 1, i.second + 1);
}